A capacitor with C = 6.00 μF is fully charged by connecting it to a battery that has emf 50.0 V. The capacitor is disconnected from the battery. A resistor of resistance R = 185 Ω is connected across the capacitor, and the capacitor discharges through the resistor.

What is the charge q on the capacitor when the current in the resistor is 0.180 A?

1.99*10^-4coulombs

Explanation:

The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor (C) is the proportionality constant. Mathematically, Q = CV

If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.

We need to calculate the voltage on the resistor first when 0.18A current is passed through it.

V = 0.18 * 185

V = 33.3Volts

The charge Q on the resistor will be;

Q = CV

Were C = 6.00 μF, V = 33.3

Q = 6*10^-6 * 33.3

Q = 0.0001998

Q = 1.99*10^-4Coulombs